35  The Electromagnetic Spectrum

35.1 Syllabus inquiry question

  • What is light?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 26:

Maxwell’s equations revealed something astounding: light is an electromagnetic wave. Radio waves, microwaves, X-rays—they’re all the same thing, just at different frequencies. Nature uses one mechanism to produce an infinite spectrum.

35.2 Learning Objectives

  • Describe Maxwell’s prediction of electromagnetic waves
  • Apply the wave equation: \(c = f\lambda\)
  • Explain historical methods to measure the speed of light
  • Use spectroscopy to identify elements and determine stellar properties
  • Apply Wien’s displacement law to stellar temperatures

35.3 Content

35.3.1 Maxwell’s Electromagnetic Theory

James Clerk Maxwell unified electricity and magnetism into a single theory (1865):

  1. Changing electric fields create magnetic fields
  2. Changing magnetic fields create electric fields
  3. These oscillations can propagate through space as waves

Maxwell calculated the speed of these waves:

\[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.00 \times 10^8\ \text{m/s}\]

This matched the measured speed of light—proving that light is an electromagnetic wave.

35.3.2 The Wave Equation

All electromagnetic waves obey:

\[c = f\lambda\]

where: - \(c = 3.00 \times 10^8\) m/s (in vacuum) - \(f\) = frequency (Hz) - \(\lambda\) = wavelength (m)

Universal Speed

All EM waves travel at the same speed in vacuum. Higher frequency means shorter wavelength.

35.3.3 Interactive: Electromagnetic Spectrum

Region Wavelength Range Frequency Range
Radio > 1 m < 300 MHz
Microwave 1 mm – 1 m 300 MHz – 300 GHz
Infrared 700 nm – 1 mm 300 GHz – 430 THz
Visible 400 – 700 nm 430 – 750 THz
Ultraviolet 10 – 400 nm 750 THz – 30 PHz
X-ray 0.01 – 10 nm 30 PHz – 30 EHz
Gamma < 0.01 nm > 30 EHz

35.3.4 Measuring the Speed of Light

Roemer (1676): Used timing variations of Jupiter’s moons to estimate c ≈ 2.2 × 10⁸ m/s

Fizeau (1849): Reflected light off a mirror 8.6 km away through a rotating toothed wheel. Got c ≈ 3.13 × 10⁸ m/s

Foucault (1862): Used rotating mirrors for more precision. Got c ≈ 2.98 × 10⁸ m/s

Modern value: \(c = 299,792,458\) m/s (exact, defines the metre)

35.3.5 Spectroscopy

Spectroscopy is the study of light spectra to identify materials and their properties.

Types of spectra: - Continuous spectrum: Hot dense object (black body) - Emission spectrum: Hot low-density gas (bright lines) - Absorption spectrum: Cool gas in front of hot source (dark lines)

Element Identification

Each element has a unique pattern of spectral lines—like a fingerprint. This allows identification of elements in distant stars.

35.3.6 Interactive: Spectral Lines

35.3.7 Wien’s Displacement Law

The peak wavelength of a black body spectrum depends on temperature:

\[\lambda_{max} = \frac{b}{T}\]

where: - \(\lambda_{max}\) = peak wavelength (m) - \(b = 2.90 \times 10^{-3}\) m·K (Wien’s constant) - \(T\) = absolute temperature (K)

Temperature and Colour
  • Hot stars (T > 10,000 K) → peak in UV/blue → appear blue-white
  • Medium stars (T ≈ 6000 K) → peak in visible → appear yellow
  • Cool stars (T < 4000 K) → peak in red/IR → appear red

35.3.8 Doppler Effect for Light

When a source moves relative to an observer, wavelengths shift:

\[\frac{\Delta\lambda}{\lambda} = \frac{v}{c}\]

  • Redshift: Source moving away → wavelength increases
  • Blueshift: Source approaching → wavelength decreases

This allows measurement of stellar velocities.

35.4 Worked Examples

35.4.1 Example 1: Wavelength from frequency

Find the wavelength of yellow light with frequency \(5.0 \times 10^{14}\) Hz.

Solution:

  1. Use \(c = f\lambda\), so \(\lambda = c/f\)

  2. \(\lambda = (3.0 \times 10^8)/(5.0 \times 10^{14})\)

  3. \(\lambda = 6.0 \times 10^{-7}\ \text{m} = 600\ \text{nm}\) (orange-yellow)

35.4.2 Example 2: Wien’s law for the Sun

The Sun has surface temperature 5800 K. Find its peak wavelength.

Solution:

  1. Use \(\lambda_{max} = b/T\)

  2. \(\lambda_{max} = (2.90 \times 10^{-3})/5800\)

  3. \(\lambda_{max} = 5.0 \times 10^{-7}\ \text{m} = 500\ \text{nm}\) (green, in visible)

35.4.3 Example 3: Stellar velocity from Doppler shift

A hydrogen spectral line at 656.3 nm is observed at 656.7 nm. Find the star’s velocity.

Solution:

  1. \(\Delta\lambda = 656.7 - 656.3 = 0.4\ \text{nm}\)

  2. \(v = c \times \Delta\lambda/\lambda = (3.0 \times 10^8) \times (0.4/656.3)\)

  3. \(v = 1.8 \times 10^5\ \text{m/s} = 180\ \text{km/s}\)

  4. Redshifted → star is moving away

35.4.4 Example 4: Fizeau’s experiment

Fizeau’s wheel had 720 teeth and rotated at 12.6 rev/s when light passed through one gap and returned through the next. The mirror was 8.6 km away. Estimate c.

Solution:

  1. Time for light to travel 2 × 8.6 km = 17.2 km

  2. Time for wheel to rotate 1/1440 revolution (half tooth spacing)

  3. \(t = \frac{1}{1440 \times 12.6} = 5.5 \times 10^{-5}\ \text{s}\)

  4. \(c = d/t = 17200/(5.5 \times 10^{-5}) = 3.1 \times 10^8\ \text{m/s}\)

35.5 Common Misconceptions

Common Misconceptions
  • Misconception: Different colours of light travel at different speeds in vacuum. Correction: All EM radiation travels at exactly c in vacuum. Speed differences only occur in materials (dispersion).

  • Misconception: Radio waves are different from light waves. Correction: They’re the same phenomenon—electromagnetic waves—just at different frequencies.

  • Misconception: Absorption spectra are created by hot gases. Correction: Absorption occurs when cool gas absorbs specific wavelengths from a continuous spectrum passing through it.

  • Misconception: Wien’s law gives the total energy radiated. Correction: Wien’s law gives the peak wavelength. Stefan-Boltzmann law (\(P \propto T^4\)) gives total power.

  • Misconception: Redshift always means the object is moving away. Correction: Gravitational redshift and cosmological expansion also cause redshift without relative motion.

35.6 Practice Questions

35.6.1 Easy (2 marks)

Calculate the frequency of red light with wavelength 700 nm.

  • Use \(f = c/\lambda\) (1)
  • \(f = (3.0 \times 10^8)/(7.0 \times 10^{-7}) = 4.3 \times 10^{14}\) Hz (1)

Answer: \(4.3 \times 10^{14}\) Hz

35.6.2 Medium (4 marks)

A star has surface temperature 4000 K. Calculate its peak wavelength and state what colour the star appears.

  • Use \(\lambda_{max} = b/T\) (1)
  • \(\lambda_{max} = (2.90 \times 10^{-3})/4000 = 7.25 \times 10^{-7}\) m (1)
  • λ = 725 nm is in infrared/red region (1)
  • Star appears red/orange (1)

Answer: \(\lambda_{max} = 725\) nm; star appears red

35.6.3 Hard (5 marks)

An astronomer observes a hydrogen line normally at 486.1 nm shifted to 486.5 nm. Calculate the star’s radial velocity and determine whether it is approaching or receding. Explain how spectral line width could indicate the star’s rotation.

  • \(\Delta\lambda = 0.4\) nm (1)
  • \(v = c \times \Delta\lambda/\lambda = (3.0 \times 10^8) \times (0.4/486.1) = 2.5 \times 10^5\) m/s (1)
  • Redshifted (longer wavelength) → receding (1)
  • Rotation causes one edge to approach, other to recede (1)
  • Results in broadened spectral lines (Doppler broadening) (1)

Answer: v = 250 km/s, receding; rotation causes line broadening as one side approaches while the other recedes.

35.7 Multiple Choice Questions

Test your understanding with these interactive questions:

35.8 Summary

Key Takeaways
  • Maxwell unified E&M and predicted c = 3.00 × 10⁸ m/s
  • Wave equation: \(c = f\lambda\)
  • All EM waves travel at c in vacuum
  • Spectroscopy: emission (bright lines), absorption (dark lines)
  • Wien’s law: \(\lambda_{max} = b/T\) (hotter = bluer)
  • Doppler shift: \(\Delta\lambda/\lambda = v/c\) (red = away, blue = toward)

35.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: